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Let $S$ be any regular CW decomposition of (a space homeomorphic to) the $n$-sphere, and consider a cell $\sigma$ of dimension $d \in \{0,\ldots,n\}$. Let $S'$ be the regular CW complex which remains after the open star of $\sigma$ is removed (that is, we remove the interior of $\sigma$ along with interiors of all cells $\tau$, if any, which contain the closure of $\sigma$ in their boundary).

I'd like to be able to make the following claim as a small part of a proof I'm writing:

$S'$ is contractible.

Is this true? It certainly seems reasonable, and I could see a straightforward proof in the case where $S$ is a finite triangulation. If $S$ has only finitely many cells to begin with, I could obtain a triangulation by barycentric subdivision and then the desired statement holds. I'd really like to avoid that, but I worry about pathological examples like the Alexander horned sphere, etc.

So if I can't make the claim above, are there reasonable hypotheses to impose on $S$ (like: only finitely many cells allowed) which will make my desired claim true?

Edit of course there are only finitely many cells, see Mincong Zeng's comment. But do we need anything stronger in case the claim doesn't hold as-is?

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    $\begingroup$ If my understanding is correct, by Prop 1 in appendix A of Hatcher, a CW complex is compact if and only if it has finitely many cells. $\endgroup$
    – Mingcong Zeng
    Commented May 29, 2015 at 20:29
  • $\begingroup$ Do you know a counterexample if you drop the assumption of regularity? (to save people time: a regular CW complex is one in which all attaching maps are homeomorphisms onto their images) $\endgroup$
    – John Pardon
    Commented May 29, 2015 at 23:47
  • $\begingroup$ @JohnPardon: Start with a square, identify the left and right edges to a single edge $\sigma$ so you have a tube, then attach disks as top and bottom caps to make the boundary of a cylinder (homeomorphic to $S^2$). The open star of $\sigma$ is the complement of the disjoint top and bottom disks. $\endgroup$
    – Tyler Lawson
    Commented May 30, 2015 at 0:08
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    $\begingroup$ @TylerLawson Or you could use the minimal CW circle that glues the boundary of an interval to a single vertex. Then the open star of that vertex is everything and its complement is therefore empty. $\endgroup$
    – Vidit Nanda
    Commented May 30, 2015 at 0:18
  • $\begingroup$ @ViditNanda Great, that's much simpler. $\endgroup$
    – Tyler Lawson
    Commented May 30, 2015 at 2:26

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